Ex 12.2, 10
There are two types of fertilisers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs 6/kg and F2 costs
Rs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Let the fertilizer F1 requirement be x kg
fertilizer F2 requirement be y kg
According to Question :
As we need to Minimize the cost,
hence the function used here is Minimize z.
Cost of fertilizer F1 = Rs 6 k/g
Cost of fertilizer F2 = Rs 5 k/g
Minimize z = 6x + 5y
Combining all Constraints :
Min Z = 6x + 5y
Subject to constraints
2x + y ≥ 280
3x + 5y ≥ 700
x, y ≥ 0
As the feasible area is unbounded
Hence 1000 may or may not be the minimum value of Z.
For this, we need to graph inequality
6x + 5y < 1000
Since, there is no common point between the feasible region and 6x + 5y < 1000
Hence the cost will be minimum, if
Fertilizer F1 used = 100 kg
Fertilizer F2 used = 80 kg
Minimum Cost = Rs 1000